Problem: A sphere intersects the $xy$-plane in a circle centered at $(2,4,0)$ with radius 1.  The sphere also intersects the $yz$-plane in a circle centered at $(0,4,-7),$ with radius $r.$  Find $r.$
Answer: The center of the sphere must have the same $x$- and $y$-coordinates of $(2,4,0).$  It must also have the same $y$- and $z$-coordinates as $(0,4,-7).$  Therefore, the center of the sphere is $(2,4,-7).$

[asy]
import three;

size(250);
currentprojection = perspective(6,3,2);

real t;
triple P, Q;

P = (2,4,0) + (Cos(330),Sin(330),0);
Q = (0,4,-7) + sqrt(46)*(0,Cos(0),Sin(0));

path3 circ = (0,4 + sqrt(46),-7);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((0,4,-7) + sqrt(46)*(0,cos(t),sin(t)));
}

draw(surface(circ--cycle),palecyan,nolight);
draw(circ,red);

circ = (3,4,0);
for (t = 0; t <= 2*pi + 0.1; t = t + 0.1) {
  circ = circ--((2,4,0) + (cos(t),sin(t),0));
}

draw(surface(circ--cycle),paleyellow,nolight);
draw(circ,red);

draw((5,0,0)--(-1,0,0));
draw((0,12,0)--(0,-1,0));
draw((0,0,-14)--(0,0,1));
draw(P--(2,4,0)--(2,4,-7)--(0,4,-7));
draw(P--(2,4,-7)--Q--(0,4,-7));

dot("$(2,4,0)$", (2,4,0), N);
dot("$(0,4,-7)$", (0,4,-7), NE);
dot("$(2,4,-7)$", (2,4,-7), S);
dot("$P$", P, SW);
dot("$Q$", Q, E);

label("$x$", (5.2,0,0), SW);
label("$y$", (0,12.2,0), E);
label("$z$", (0,0,1.2), N);
label("$1$", (P + (2,4,0))/2, SE);
label("$7$", (2,4,-3.5), E);
label("$2$", (1,4,-7), NW);
label("$r$", (Q + (0,4,-7))/2, NE);
[/asy]

Let $P$ be a point on the circle centered at $(2,4,0)$ with radius 1.  Then $P,$ $(2,4,0),$ and $(2,4,-7)$ form a right triangle, which tells us that the radius of the sphere is $\sqrt{1^2 + 7^2} = 5 \sqrt{2}.$

Let $Q$ be a point on the circle centered at $(0,4,-7)$ with radius $r.$  Then $Q,$ $(0,4,-7),$ and $(2,4,-7)$ form a right triangle, which tells us that the $r = \sqrt{50 - 2^2} = \boxed{\sqrt{46}}.$